Danelectro Fab Tone "Bass Mod"?

Discussion in 'The Projects Forum' started by Dash'sLashes, May 2, 2014.

  1. Dash'sLashes

    Thread Starter New Member

    May 2, 2014
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    [​IMG]
    Hey! I would like to ask about modding a Danelectro Fab Tone to better work with my bass! All I can find online is i heard you change out C5 from a 10n to something bigger, im going to put a .47uF into C5 and change the clipping diodes to those of a klon I guess that lets more lows thu(?). Can anyone here recommend any other mods that will help me use this great sounding distortion for BASS!?

    Schematic: http://www.geocities.ws/diygescorp/danelectrofabtone.gif
     
  2. Brevor

    Active Member

    Apr 9, 2011
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    Thats a crazy drawn schematic, you may try jumping C4 or C9 with a larger value and see if the bass response improves. You dont need to remove them just solder another cap across them on the bottom of the board as a test, if you like the result then you can replace the original cap with the larger one.
     
  3. Dash'sLashes

    Thread Starter New Member

    May 2, 2014
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    thanks for taking a look for me! So far, I have plans to switch C4 and C9 to from .33 to 1uF, and C5 from 10n to .47uF... and C19 from .33 to 1uF tantalum. As I mentioned, switching to nice Ge Diodes too. I want to drop in some switchcraft 1/4" jacks somehow if they'll fit in the box. Am I missing anything else on the circuit to successfully rock bass thu her though?
     
  4. #12

    Expert

    Nov 30, 2010
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    Not a fun job, but here's how I see it:

    C102 = OK
    C1 = OK
    C4 = marginal @ 48Hz
    C5 = Inverse function. Making it larger will make the bass end worse.
    C6 = unknown. Can't understand it. Try something 2X or 10X the original size and see what happens.
    C9 = marginal @ 48 Hz
    C10 = Marginal @ 33 Hz
    C13 = OK
    C14, C15 = not OK @ 72 Hz. Try at least 4X the original size. Change them as a pair.

    I'll be amazed if you don't find a mistake in this list.
     
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  5. Dash'sLashes

    Thread Starter New Member

    May 2, 2014
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    thanks for looking at it! your reccomendation for C5 is my only conflict. I've heard a couple people say to put a jumper in C5 and bypass that cap altogether. Why do you say to leave it alone? What will happen with the .47uF in it? Why do people say jump it? Im not really really smart, but i thought I'd put a .47uF mustard into C5 for tone. am I off the deep end???
     
  6. MrChips

    Moderator

    Oct 2, 2009
    12,415
    3,354
    I would not jumper across C5. Doing so will disturb the DC bias voltages on the transistor.

    I would experiment with placing a larger capacitor (0.1μF - 1μF) across C4 and C9.
     
  7. #12

    Expert

    Nov 30, 2010
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    C5: I didn't tell you to leave it alone. I said it works backwards to what you'd expect. You're welcome to play with it all you want. Let us know how it turns out.
     
  8. Dash'sLashes

    Thread Starter New Member

    May 2, 2014
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    ok! so... the input and output caps wee all know should be up-ed. the sticker is C5. I have two nice vintage mullard mustards. One is .1uF, and the other is .47uF, ill try them both so wee'll see! thanks you guys. Still open for other ideas too. paid $10 for this pedal and it sounds like a $400 distortion imo.
     
  9. #12

    Expert

    Nov 30, 2010
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    Just for exercise, let's examine this tone stage.
    Audio frequencies arrive through C4 and are DC corrected with R6.
    The job of IC1a is to make the voltage at it's inverting input equal to the voltage at its non-inverting input.
    The gain equation is: Vo = Vin x (1+ Rf/Rc)
    Rf is 220k in parallel with 100 pf. That pair sets a frequency change at 7234 Hz.
    That means, the currents through C7 and R10 are equal to each other at 7234 Hz.
    And that means frequencies above 7234 Hz have more effect on the inverting input than lower frequencies, and, this being the inverting input, those higher frequencies will be diminished at the output of IC1a compared to lower frequencies.
    Bingo. We have high limit.

    Now, the rest of the circuit is defined as Rc.
    The inverting input has the same signal as what came in through C4 because that is the job of IC1a. It will do whatever is necessary at its output to get the two inputs to be equal.
    C5 and C6 feed higher frequencies more easily than they feed lower frequencies. Q4 is an emitter follower, so it has no AC gain and lowers the impedance of the signal. This low impedance signal arrives at the emitter of Q4, in phase, and is redirected back to the junction of C5 and C6.

    Here's where it gets sticky for me.
    If C5 and C6 are calculated in series, they are equal to 6.875n
    6.875n in series with 47k makes a circuit with a frequency change point of 493 Hz. That sounds reasonable for our frequency range, but it isn't the truth about how the circuit works.
    Whatever passes through C5 (considering its relationship with 47k) gets fed back into the circuit at the junction of C5 & C6. Same phase and same amplitude. DC is irrelevant because it is isolated between 2 capacitors. So, I will substitute an imaginary gain block of +1 to simplify the circuit. That means there will be no AC voltage across C5. (R7 is represented as going to ground because it is going to an AC ground in reality.) This simplifies the AC current as going through C6 and R7.

    At 100 Hz, C6 is 72.3 K ohms and C5 is 159 k ohms.
    At 339 Hz, C6 is 21.4 k ohms and C5 is 47 k ohms.
    At 1000 Hz, C6 is 7.23 k ohms and C5 is 15.9k ohms.

    In the 100 Hz circuit, the voltage at the junction of C6 and C5 is 39% of the voltage at the inverting input of IC1a.
    In the 339 Hz circuit, the voltage at the junction of C6 and C5 is 69% of the voltage at the inverting input of IC1a.
    In the 1000 Hz circuit, the voltage at the junction of C6 and C5 is 87% of the voltage at the inverting input of IC1a.

    The higher the frequency, the more voltage is at the C6-C5 junction.
    Therefore the higher the frequency, the more impedance Rc has.
    Therefore, in accordance with gain = 1+ (Rf/Rc) the higher Rc is, the lower the gain of IC1a is.

    Now that we have a working model, we can see what happens if you increase C6.
    Change C6 from 22n to 220n and its impedance goes down by a factor of 10.
    The 339 Hz model becomes C6 = 2.14k in series with 47k for a voltage of 96% at the junction of C6-C5
    The remaining voltage is higher so the impedance to ground must be higher and the gain of the whole circuit becomes lower.

    Increasing C6 lowers the gain at any particular frequency.
    This is the opposite of what one normally expects in an audio circuit because the normal expectation is that making the capacitors larger increases gain.

    As usual, if anybody can see where I messed up the logic, please bring it to our attention.
    Don't bother picking apart the exactness of the resistances or impedances. That wasn't the purpose of the exercise.
     
    Last edited: May 4, 2014
  10. Dash'sLashes

    Thread Starter New Member

    May 2, 2014
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    My skills are basic and as follows: Ask someone on the internet what capacitors to change and ask them what values will make the pedal work best for bass.

    It sounds like I can hear the pedal cut the bass in my signal, like shelving around 70Hz and below. I can NOT hear a very strong bass guitar fundamental playing a detuned low C# when the pedal is engaged which is about in the 40Hzs to the best of my knowledge. im just a punk rocker and bass pedal tinkerer. I really cant do math problems more than a few times a few or deviding on a calculator. I learned a few things in your write-up thanks! SO my plan right now is to change the following:

    first:
    C4 and C9 to from .33 to 1uF
    C19 - from .33 to 1uF tantalum
    C14 and C15 - from .33 to 1uF

    then:
    C5 - try a .47uF and see, and then try a .1uF and see.

    Thant was funny when you said "I didn't tell you to leave it alone" hahah :)
     
  11. #12

    Expert

    Nov 30, 2010
    16,248
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    I mostly did that because I needed it. That tone circuit is still not completely clear to me!
    I can see which way the frequency change is going, but I can't imagine how some designer came up with an emitter follower in that circuit. Never seen it before, don't have enough math to understand it.

    The good part is that you can change capacitors and see what happens without knowing the math.
     
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